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### CMSC 56 | Lecture 9: Functions Representations

• 1. Functions Representations Lecture 9, CMSC 56 Allyn Joy D. Calcaben
• 3. Definition 1 Let A and B be two sets. A function f from A to B, denoted f : A → B , is an assignment of exactly one element of B to each element of A. We write f(a) = b if b is the unique element of B assigned by the function f to the element a of A. Functions
• 4. Definition 1 Let A and B be two sets. A function f from A to B, denoted f : A → B , is an assignment of exactly one element of B to each element of A. We write f(a) = b if b is the unique element of B assigned by the function f to the element a of A. Functions are sometimes also called mappings or transformations Functions
• 5. A f: A → B B Functions
• 6. A f: A → B B Functions X NOT ALLOWED !!
• 7. Definition 2 If f is a function from A to B, we say that A is the domain of f and B is the codomain of f. If f(a) = b, we say that b is the image of a and a is the preimage of b. The range, or image, of f is the set of all images of elements of A. Also, if f is a function from A to B, we say that f maps A to B. Functions
• 8. What are the domain, codomain, and range of the function that assigns grades to students in a discrete mathematics class. Suppose that the grades are 1.00 for Ronn, 3.00 for Ricca, 1.75 for Robien, 2.50 for Zin, 3.00 for Angelique, and 1.50 for Mace. Example
• 9. If f is a function specified by R, then f(x) is the grade of x, where x is a student. Solution
• 10. If f is a function specified by R, then f(x) is the grade of x, where x is a student. f(Ronn) = 1.00 f(Ricca) = 3.00 f(Robien) = 1.75 f(Zin) = 2.50 f(Mace) = 1.50 f(Angelique) = 3.00 Solution
• 11. If f is a function specified by R, then f(x) is the grade of x, where x is a student. f(Ronn) = 1.00 Ronn • f(Ricca) = 3.00 Ricca • f(Robien) = 1.75 Robien • f(Zin) = 2.50 Zin • f(Mace) = 1.50 Mace • f(Angelique) = 3.00 Angelique • Solution • 1.00 • 1.25 • 1.50 • 1.75 • 2.00 • 2.25 • 2.50 • 2.75 • 3.00
• 12. Set A (Domain) {Ronn, Ricca, Robien, Zin, Angelique, Mace} Set B (Codomain) {1.00, 1.25, 1.50, 1.75, 2.00, 2.25, 2.50, 2.75, 3.0} Range {1.00, 1.50, 1.75, 2.50, 3.0} Solution
• 13. Let R be the relation with ordered pairs (Chloei, 21), (Ezekiel, 22), (Hannah, 24), (Jasper, 21), (Jayven, 23), (Kent, 22), (Maren, 23), (Robyn, 24), (Sean, 22), and (Vinze, 25). Here each pair consists of a graduate student and their age. Specify a function determined by this relation. Challenge (1/4 Sheet Paper)
• 14. If f is a function specified by R, then f(x) is the age of x, where x is a student. Solution
• 15. If f is a function specified by R, then f(x) is the age of x, where x is a student. f(Chloei) = 21 f(Ezekiel) = 22 f(Hannah) = 24 f(Jasper) = 21 f(Jayven) = 23 f(Kent) = 22 f(Maren) = 23 f(Robyn) = 24 f(Sean) = 22 f(Vinze) = 25. Solution
• 16. If f is a function specified by R, then f(x) is the age of x, where x is a student. Set A (Domain) = { Chloei, Ezekiel, Hannah, Jasper, Jayven, Kent, Maren, Robyn, Sean, Vinze } Solution
• 17. If f is a function specified by R, then f(x) is the age of x, where x is a student. Set A (Domain) = { Chloei, Ezekiel, Hannah, Jasper, Jayven, Kent, Maren, Robyn, Sean, Vinze } Set B (Codomain) = { Z+ | f(x) < 100 } Solution
• 18. If f is a function specified by R, then f(x) is the age of x, where x is a student. Set A (Domain) = { Chloei, Ezekiel, Hannah, Jasper, Jayven, Kent, Maren, Robyn, Sean, Vinze } Set B (Codomain) = { Z+ | f(x) < 100 } Range = { 21, 22, 23, 24, 25 } Solution
• 19. Definition 3 If f1 and f2 be functions from A to R. Then f1 + f2 and f1 f2 are also functions from A to R defined for all x ϵ A by ( f1 + f2 )(x) = f1(x) + f2 (x) ( f1 f2 )(x) = f1(x) f2 (x) Functions
• 20. Let f1 and f2 be functions from R to R such that f1(x) = x2 and f2(x) = x – x2. What are the functions f1 + f2 and f1 f2 ? Example
• 21. Let f1 and f2 be functions from R to R such that f1(x) = x2 and f2(x) = x – x2. What are the functions f1 + f2 and f1 f2 ? ( f1 + f2 )(x) = f1(x) + f2 (x) = x2 + (x – x2) = x Solution
• 22. Let f1 and f2 be functions from R to R such that f1(x) = x2 and f2(x) = x – x2. What are the functions f1 + f2 and f1 f2 ? ( f1 + f2 )(x) = f1(x) + f2 (x) = x2 + (x – x2) = x ( f1 f2 )(x) = f1(x) f2 (x) = x2 (x – x2) = x3 – x4 Solution
• 23. Definition 4 Let f be a function from A to B and let S be a subset of A. The image of S under the function f is the subset of B that consists of the images of the elements of S. We denote the image of S by f(S), so f(S) = { t | Ǝs ϵ S ( t = f(s) ) } Functions
• 24. Definition 4 Let f be a function from A to B and let S be a subset of A. The image of S under the function f is the subset of B that consists of the images of the elements of S. We denote the image of S by f(S), so f(S) = { t | Ǝs ϵ S ( t = f(s) ) } We also use the shorthand { f(s) | s ϵ S } to denote this set. Functions
• 26. Definition 5 A function f is said to be one-to-one, or an injunction, if and only if f(a) = f(b) implies that a = b for all a and b in the domain of f. A function is said to be injective if it is one-to-one. Injuction Functions
• 27. Definition 5 A function f is said to be one-to-one, or an injunction, if and only if f(a) = f(b) implies that a = b for all a and b in the domain of f. A function is said to be injective if it is one-to-one. We can express that f is one-to-one using quantifiers as ꓯaꓯb( f(a) = f(b) → a = b ) or equivalent ꓯaꓯb( a ≠ b → f(a) ≠ f(b) ), where the universe of discourse is the domain of the function. Injuction Functions
• 28. Determine whether the function f from { a, b, c, d } to { 1, 2, 3, 4, 5 } with f(a) = 4, f(b) = 5, f(c) = 1, and f(d) = 3 is one-to-one. Example
• 29. Determine whether the function f from { a, b, c, d } to { 1, 2, 3, 4, 5 } with f(a) = 4, f(b) = 5, f(c) = 1, and f(d) = 3 is one-to-one. a • • 1 b • • 2 c • • 3 d • • 4 • 5 Solution
• 30. Determine whether the function f from { a, b, c, d } to { 1, 2, 3, 4, 5 } with f(a) = 4, f(b) = 5, f(c) = 1, and f(d) = 3 is one-to-one. a • • 1 b • • 2 c • • 3 d • • 4 • 5 Solution The function f is one-to-one because f takes on different values at the four elements of its domain.
• 31. Definition 6 A function f whose domain and codomain are subsets of the set of real numbers is called increasing if f(x) ≤ f(y), and strictly increasing if f(x) < f(y), whenever x < y and x and y are in the domain of f. Increasing & Decreasing Functions
• 32. Definition 6 Similarly, f is called decreasing if f(x) ≥ f(y), and strictly decreasing if f(x) > f(y), whenever x < y and x and y are in the domain of f. Increasing & Decreasing Functions
• 33. Definition 6 Similarly, f is called decreasing if f(x) ≥ f(y), and strictly decreasing if f(x) > f(y), whenever x < y and x and y are in the domain of f. NOTE: Strictly increasing and strictly decreasing functions are one-to-one. Increasing & Decreasing Functions
• 34. Definition 7 A function f from A to B is called onto, or a surjection, if and only if for every element b ϵ B there is an element a ϵ A with f(a) = b. A function f is called surjective if it is onto. Surjection Functions
• 35. Definition 7 A function f from A to B is called onto, or a surjection, if and only if for every element b ϵ B there is an element a ϵ A with f(a) = b. A function f is called surjective if it is onto. Remark: A function f is onto if ∀y∃x( f(x) = y ), where the domain for x is the domain of the function and the domain for y is the codomain of the function. Surjection Functions
• 36. Let f be the function from { a, b, c, d } to { 1, 2, 3 } defined by f(a) = 3, f(b) = 2, f(c) = 1, and f(d) = 3. Is f an onto function? Example
• 37. Let f be the function from { a, b, c, d } to { 1, 2, 3 } defined by f(a) = 3, f(b) = 2, f(c) = 1, and f(d) = 3. Is f an onto function? a • • 1 b • • 2 c • • 3 d • Solution
• 38. Let f be the function from { a, b, c, d } to { 1, 2, 3 } defined by f(a) = 3, f(b) = 2, f(c) = 1, and f(d) = 3. Is f an onto function? a • • 1 b • • 2 c • • 3 d • Solution Since all three elements of the codomain are images of elements in the domain, we see that f is onto.
• 39. Definition 8 The function f is a one-to-one correspondence, or a bijection, if it is both one-to-one and onto. We also say that such a function is bijective. Bijective Functions
• 40. Let f be the function from { a, b, c, d } to { 1, 2, 3, 4 } with f(a) = 4, f(b) = 2, f(c) = 1, and f(d) = 3. Is f a bijection? Example
• 41. Let f be the function from { a, b, c, d } to { 1, 2, 3, 4 } with f(a) = 4, f(b) = 2, f(c) = 1, and f(d) = 3. Is f a bijection? a • • 1 b • • 2 c • • 3 d • • 4 Example
• 42. Let f be the function from { a, b, c, d } to { 1, 2, 3, 4 } with f(a) = 4, f(b) = 2, f(c) = 1, and f(d) = 3. Is f a bijection? a • • 1 b • • 2 c • • 3 d • • 4 Example Since the function f is one-to-one and onto, then function f is a bisection.
• 43.
• 44. Let A be a set. The identity function on A is the function ιA : A → A, where ιA(x) = x for all x ϵ A. In other words, the identity function ιA is the function that assigns each element to itself. The function ιA is one-to-one and onto, so it is a bijection. Identity Function
• 45. Let A = { 1, 2, 3 }, what is ιA(1)? ιA(2)? ιA(3)? Example
• 46. Let A = { 1, 2, 3 }, what is ιA(1)? ιA(2)? ιA(3)? ιA(1) = 1 ιA(2) = 2 ιA(3) = 3 Solution
• 47.
• 49. Definition 9 Let f be a one-to-one correspondence from the set A to the set B. The inverse function of f is the function that assigns to an element b belonging to B the unique element a in A such that f(a) = b. The inverse function of f is denoted by f -1. Hence, f -1(b) = a when f(a) = b. Inverse Functions
• 50. Be sure not to confuse the function f -1 with the function 1 / f , which is the function that assigns to each x in the domain the value 1 / f(x). Notice that the latter makes sense only when f(x) is a non-zero real number. Inverse Functions
• 51. Be sure not to confuse the function f -1 with the function 1 / f , which is the function that assigns to each x in the domain the value 1 / f(x). Notice that the latter makes sense only when f(x) is a non-zero real number. A one-to-one correspondence is called invertible because we can define an inverse of this function. A function is not invertible if it is not a one- to-one correspondence, because the inverse of such a function does not exist. Inverse Functions
• 52. Let f be the function from { a, b, c } to { 1, 2, 3 } such that f(a) = 2, f(b) = 3, and f(c) = 1. Is f invertible, and if it is, what is its inverse? Example
• 53. Let f be the function from { a, b, c } to { 1, 2, 3 } such that f(a) = 2, f(b) = 3, and f(c) = 1. Is f invertible, and if it is, what is its inverse? a • • 1 b • • 2 c • • 3 Solution
• 54. Let f be the function from { a, b, c } to { 1, 2, 3 } such that f(a) = 2, f(b) = 3, and f(c) = 1. Is f invertible, and if it is, what is its inverse? a • • 1 f -1(1) = c b • • 2 f -1(2) = a c • • 3 f -1(3) = b Solution
• 55. Let f be the function from { a, b, c } to { 1, 2, 3 } such that f(a) = 2, f(b) = 3, and f(c) = 1. Is f invertible, and if it is, what is its inverse? a • • 1 f -1(1) = c b • • 2 f -1(2) = a c • • 3 f -1(3) = b Solution The function f is invertible because it is a one-to-one correspondence.
• 56. Let f be the function from R to R with f(x) = x2. Is f invertible? Example
• 57. Let f be the function from R to R with f(x) = x2. Is f invertible? Because f(-2) = f(2) = 4, f is not one-to-one. If an inverse function were defined, it would have to assign two elements to 4. Hence, f is not invertible. Solution
• 59. Definition 10 Let g be a function from the set A to the set B and let f be a function from the set B to the set C. The composition of the functions f and g, denoted for all a ϵ A by f ◦ g, is defined by (f ◦ g)(a) = f(g(a)) Composition of Functions
• 60.
• 61. Let g be the function from the set { a, b, c } to itself such that g(a) = b, g(b) = c, and g(c) = a. Let f be the function from the set { a, b, c } to the set { 1, 2, 3 } such that f(a) = 3, f(b) = 2, and f(c) = 1. What is the composition of f and g, and what is the composition of g and f? Example
• 62. Let g be the function from the set { a, b, c } to itself such that g(a) = b, g(b) = c, and g(c) = a. Let f be the function from the set { a, b, c } to the set { 1, 2, 3 } such that f(a) = 3, f(b) = 2, and f(c) = 1. What is the composition of f and g, and what is the composition of g and f? Composition of f and g: (f ◦ g)(a) = f(g(a)) = f(b) = 2 (f ◦ g)(b) = f(g(b)) = f(c) = 1 (f ◦ g)(c) = f(g(c)) = f(a) = 3 Solution
• 63. Let g be the function from the set { a, b, c } to itself such that g(a) = b, g(b) = c, and g(c) = a. Let f be the function from the set { a, b, c } to the set { 1, 2, 3 } such that f(a) = 3, f(b) = 2, and f(c) = 1. What is the composition of f and g, and what is the composition of g and f? Composition of f and g: Composition of g and f: (f ◦ g)(a) = f(g(a)) = f(b) = 2 (f ◦ g)(b) = f(g(b)) = f(c) = 1 (f ◦ g)(c) = f(g(c)) = f(a) = 3 Solution g ◦ f is not defined, because the range of f is not a subset of the domain g.
• 64. Let f and g be the functions from the set of integers to the set of integers deﬁned by f(x) = 2x + 3 and g(x) = 3x + 2. What is the composition of f and g? What is the composition of g and f? Challenge (1/4 Sheet Paper)
• 65. Let f and g be the functions from the set of integers to the set of integers deﬁned by f(x) = 2x + 3 and g(x) = 3x + 2. What is the composition of f and g? What is the composition of g and f? Composition of f and g: (f ◦ g)(x) = f(g(x)) = f(3x + 2) = 2(3x + 2) + 3 = 6x + 7 Composition of g and f: (g ◦ f)(x) = g(f(x)) = g(2x + 3) = 3(2x + 3) + 2 = 6x + 11 Solution
• 68. 46 37 42 2 35 7 A B C D E F 88.85% 70.52% 82.92% 28.13% 71.88% 36.11% A B C D E F Passed,… Failed, 17.02% Best Student Performance with only 11.15% FAILURE Lowest Student Performance with almost 72% FAILURE CMSC 56 1st Long Exam| AY 2018 – 2019 EXAM PARTSHistogram Students passed in Part A 08Students failed 39Students passed Students failed in Part D 46 45 PARTA PARTD
• 69. ANSWER KEYON SELECTED PARTS PARTD 1. Thereissomeoneinthis classwhodoesnothavea goodattitude. x = “studentinthisclass” A(x)= “x hasagoodattitude” a. Ǝx ~A(x)OriginalStatement b. ~(Ǝx~A(x))= ꓯxA(x) “Everyoneinthisclasshasa goodattitude.” 2. Norabbitknows Calculus. x = “rabbit” C(x)= “x knowscalculus” a.~ƎxC(x)OriginalStatement b. ~(~ƎxC(x)) = Ǝx A(x) “Thereisa rabbitthat knows Calculus.” PARTF 1. I driveifit istoo dangerousortoo fartobike. p = “Idrive” q = “Itistoo dangerous” r = “Itistoo fartobike” (qV r)→ p OriginalStatement a. p → (qV r)Converse “IfI drive,thenit istoo dangerousor too fartobike.” “Itistoo dangerousortoo fartobike ifI drive.” b. ~p→~(qV r)Contrapositive “IfI don’t drive,thenitisnottoo dangerousortoo fartobike.” “IfI don’t drive,thenitisnottoo dangerousandnottoofartobike.” “Itisnottoodangerousortoo farto bike ifIdon’t drive.” “Itisnottoodangerousand nottoo farto bike if I don’tdrive.” 2. a. Nostudentinyourclasshastaken acourseinlogic programming. x = “studentsinthisclass” L(x)= “x hastakena coursein logicprogramming” ꓯx~L(x) b. ComputerScienceis easyorcampingisfun, aslong asitissunnyand thehomework isdone. m =“computerscienceiseasy” c = “campingisfun” s = “itissunny” h = “homeworkisdone” (sꓥ h) ↔ ( m V c ) (sꓥ h) → ( m V c )
• 70. 100.0% Garcia, Patricia Marie 96.0% Samson, Aron Miles 93.0% Honeyman, John 91.0% Jomoc, Gracielou Patot, Oliver 90.0% Garcia, Michael John CMSC 56 1st LE Top 5 Scorers

### Editor's Notes

1. Note that the functions f1 + f2 and f1f2 have been deﬁned by specifying their values at x in terms of the values of f1 and f2 at x.
2. The notation f(S) for the image of the set S under the function f is potentially ambiguous. Here, f(S)denotes a set, and not the value of the function f for the set S.
3. Why? One-to-one function: A function is one-to-one if and only if f(x)  f(y), whenever x  y.
4. ι = Greek letter iota
5. ι = Greek letter iota
6. ι = Greek letter iota
7. Function f is not onto.
8. In other words, f ◦ g is the function that assigns to the element a of A the element assigned by f to g(a). That is, to ﬁnd (f ◦ g)(a) we ﬁrst apply the function g to a to obtain g(a) and then we apply the function f to the result g(a) to obtain (f ◦ g)(a) = f(g(a)). Note that the composition f◦ g cannot be deﬁned unless the range of g is a subset of the domain of f . In Figure 7 the composition of functions is shown.
9. Remark: Note that even though f ◦ g and g ◦ f are deﬁned for the functions f and g in Example 23, f ◦ g and g ◦ f are not equal. In other words, the commutative law does not hold for the composition of functions.
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